This question is quite unique.
Does there exist some point in the coordinate system such that any line passing through it has at most 2 rational points lying on it?
This question is quite unique.
Does there exist some point in the coordinate system such that any line passing through it has at most 2 rational points lying on it?
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The existence of 2 rational points on a line $L$ implies the existence of infinitely many. For if a line $L$ has two rational points $p=(x_1,y_2)$, $q=(x_2,y_2)$, then $L$ satisfies an equation with rational coefficients $ax+by=c$, which one can obtain by rewriting the "two-point" form of the equation for $L$ which is $(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)$. If $L$ is vertical, equivalently if $b = 0$, then clearly it contains infinitely many rational points. If $L$ is not vertical then substituting any rational value $x$, and then solving for $y$, the result is a rational value of $y$, and by this means we get infinitely many rational points on $L$.
So your question is equivalent to asking whether there exists some point $P$ such that any line passing through it has at most 1 rational point on it.
NOTE: My last attempt had an error after this point, here is a correct solution.
Take $P=(r,s)$ such that, in the vector space structure of $\mathbb{R}$ over the field $\mathbb{Q}$, the set $\{1,r,s\}$ is linearly independent. For example, $r=\sqrt{2}$, $s=\sqrt{3}$. $P$ cannot lie on a line $L$ with two rational points because if it does then, as just shown above, the line $L$ has an equation with rational coefficients $ax+by=c$. Plugging in $x=r$, $y=s$ gives $ar+bs=c$ which gives $-c1+ar+bs=0$, contradicting linear independence of $\{1,r,s\}$.